Download - Inverse substitution rule Inverse Substitution Rule If and is differentiable and invertible. Then
Inverse substitution rule Inverse Substitution Rule If and
is differentiable and invertible. Then
( ( )) ( ) ( )f g t g t dt F t C ( )x g t
1( ) ( ( )) ( ) ( ) ( ( )) .f x dx f g t g t dt F t C F g x C
Example: trigonometric substitutions Ex. Evaluate
Sol. Let then
From we derive
2 2 ( 0).a x dx a sin ,x a t cos .dx a tdt
2 2 2 2cos cos cosa x dx a t a tdt a tdt 2 2 sin 2
(1 cos 2 ) ( )2 2 2
a a tt dt t C
sin ,x a t2 2
arcsin , cos .x a x
t ta a
2 2 22 2 arcsin .
2 2
a x x a xa x dx C
a
Example: trigonometric substitutions Ex. Evaluate
Sol. Let then
From we derive
2 2( 0).
dxa
x a
tan ,x a t 2sec .dx a tdt1 1 sin
ln | |2 1 sin
tC
t
tan ,x a t2 2
sin .x
tx a
2
2 2
secsec
sec
dx a tdt tdt
a tx a
2 2 2 2 2
22 2
1 ( )ln ln .
2
dx x a x x a xC C
a ax a
Example Ex. Evaluate
Sol.
2
2
9.
xdx
x
22 2
2 2 2
9 (3sin )9 cos(3sin )
(3sin ) sin
tx tdx d t dt
x t t
22 9
(csc 1) cot arcsin .3
x xt dt t t C C
x
Example Ex. Evaluate substitution
Ex. Find substitution
Ex. Find
substitution
2 2
1.
4dx
x x 2 tanx t
2 2
1.
4dx
x x 2secx t
2
1.
3 2dx
x x x 2 23 2 4 ( 1)x x x 1 2sinx t
Example: reciprocal substitution Ex. Evaluate
Sol. Let then
2.
3 2 1
dx
x x x
1,x
t
2
1.dx dt
t
2 2 2 23 2 1 3 2 2 ( 1)
dx dt dt
x x x t t t
2
1 1 1arcsin arcsin .
2 2 211 ( )
2
dt t xC C
xt
Example: rational substitution Ex. Evaluate
Sol. Let then
.1
xdx
x2 ,x t
2 3
2 21 1
t ttdt dt
t t
3
2 21 22[ ( 1) ] 2 2ln(1 )
1 3
t
t t dt dt t t t Ct
3 1 12
1
t
dtt
322 2ln(1 ) .
3
xx x x C
2 .dx tdt
1
xdx
x
Inverse substitution for definite integral
The Inverse Substitution Rule for definite integral: If
x=g(t) is differentiable, invertible and, when x is in between
a and b, t is in between and Then1( )g a 1( ).g b
( ) ( ( )) ( ) .b
af x dx f g t g t dt
Example Ex. Evaluate
Sol. Let then and when x
changes from a to 2a, t changes from 0 to
2 22
4( 0).
a
a
x adx a
x
sec ,x a t sec tan ,dx a t t/ 3.
2 22
34 4 40
tansec tan
sec
a
a
x a a tdx a t tdt
x a t
32 33
2 2 200
1 1 3sin cos sin .
3 8t tdt t
a a a
Example Ex. Evaluate
Sol. Since let
Then and when x changes from 2 to 3, t changes from 0 to
3
2 32
1.
(4 )
xdx
x x
2 2(4 ) 4 ( 2) ,x x x 2 2sin .x t 2cos ,dx t
/ 6.3
6 63 22 32 0 0
1 3 2sin 3 2sin2cos
(2cos ) 4cos(4 )
x t tdx tdt dt
t tx x
6
0
3 1 3 3 1 7 3 6tan ( ) .
4 2cos 4 3 2 12t
t
Example: application of substitution Ex. Find
Sol. Let then
2
0
sin.
sin cos
xdx
x x
02 2
0 02
sin cos cos
sin cos cos sin sin cos
x t tdx dt dt
x x t t t t
,2
x t
2 2 2
0 0 0
sin 1 cos sin( ) .
sin cos 2 sin cos sin cos 4
x t xdx dt dx
x x t t x x
Example: application of substitution Ex. Find the definite integral
Sol.
By substitution
22
2
sin.
1 x
xdx
e
2 2 20
2 2
022
sin sin sin.
1 1 1x x x
x x xdx dx dx
e e e
2 2 2 20 0
2 2
0 02 2
sin sin sin sin.
1 1 1 1
x
x t x x
x t x e xdx dt dx dx
e e e e
,x t
2 2 222 2 2 2
0 0 02
sin sin sinsin .
1 1 1 4
x
x x x
x e x xdx dx dx xdx
e e e
Integration of rational functions Any rational function can be integrated by the following
two steps: a). express it as a sum of simpler fractions by partial fraction technique; b. integrate each partial fraction using the integration techniques we have learned.
For example, since
we have
2
5 5 2 1
2 ( 1)( 2) 1 2
x x
x x x x x x
2
5 2 1( ) 2 ln | 1| ln | 2 | .
2 1 2
xdx dx x x C
x x x x
Technique for partial fraction Take any rational function where P and Q are
polynomials. If the degree of P is less than the degree of Q, we call f a
proper fraction. If f is improper, that is, degree of P greater than or equal to
degree of Q, then (by long division) we must have
where S and R are also polynomials and degree of R less than degree of Q.
( )( ) ,
( )
P xf x
Q x
( ) ( )( ) ( ) ,
( ) ( )
P x R xf x S x
Q x Q x
Technique for partial fraction For example, by long division, we can derive
For the above reason, we only need to consider the properrational functions. The next step is to factor the denominator Q(x). It can beshown that any polynomial can be factored as a product oflinear factors (in the form ax+b) and irreducible quadraticfactors (in the form ). Forexample,
32 2
2 .1 1
x xx x
x x
2 2 with 4 0ax bx c b ac 4 2( ) -16=(x-2)(x+2)(x 4).Q x x
4 2 2 2+1 ( 1)( 1)x x x x x x
Technique for partial fraction The third step is to express the proper rational function
R(x)/Q(x) as a sum of partial fractions of the form
These two kind of rational functions can be integrated as
2 or .
( ) ( )i j
A Ax B
ax b ax bx c
11 1 1 1ln | | , ( )
( ) (1 )k
kdx ax b C dx ax b C
ax b a ax b a k
2
2 2 2
( ).
( ) ( ) [1 ( ) ]k k k
Ax B d ax bx c Fdx C D dx
ax bx c ax bx c x E
Example Ex. Find
Sol.
2
1.
( 1)n nI dx
x
2 2 2
1
( 1) ( 1) ( 1)n n n n
dx xI xd
x x x
2
2 2 12
( 1) ( 1)n n
x x dxn
x x
2
2 2 1
( 1 1)2
( 1) ( 1)n n
x x dxn
x x
122 2 .
( 1) n nn
xnI nI
x
1 2
2 1.
2 2 ( 1)n n n
n xI I
n n x
1 arctan .I x C
2 2
arctan.
2 2( 1)
x xI C
x
Technique for partial fraction From the above analysis, we see that how to split a rational
function into partial fractions is the key step to integrate the rational function.
When Q(x) contains factor the partial fractions contain
When Q(x) contains irreducible factor the partial fractions contain
( ) ,kx a
1 22
.( ) ( )
kk
AA A
x a x a x a
2( ) ,kx px q
1 1 2 22 2 2 2
.( ) ( )
k kk
B x DB x D B x D
x px q x px q x px q
Example Ex. Find
Sol. Since the partial fraction has the form
Expanding the right side and comparing with the left side,
2
3 2
2 1.
2 3 2
x xdx
x x x
3 22 3 2 ( 2)(2 1),x x x x x x
2
3 2
2 1.
2 3 2 2 2 1
x x A B C
x x x x x x
2 2 1 ( 2)(2 1) (2 1) ( 2). x x A x x Bx x Cx x2(2 2 ) (3 2 ) 2 A B C x A B C x A
1 1 1, ,
2 5 10 A B C
Example Ex. Find Sol.
5 3 2
4 2
2 7.
2 2
x x x x
I dxx x x
5 3 2
4 2 4 2
2 7 5
2 2 2 2
x x x x x
xx x x x x x
4 2 2 22 2 ( 1) ( 2 2) x x x x x x
1 24 2 2 2
5
2 2 1 ( 1) 2 2
A Ax Bx D
x x x x x x x
1 2
1 1 8, 1, ,
5 5 5 A A B D
221 1 1 7
ln | 1| ln( 2 2) arctan( 1) .2 5 1 10 5
xI x x x x C
x
Remark There are two methods to find the coefficients in the
partial fractions. One method is comparing the corresponding coefficients of polynomials on both sides; the other is taking some special values of x in the identity.
For instance, in the last example, we have2 2 2
1 25 ( 1)( 2 2) ( 2 2) ( )( 2 1) x A x x x A x x Bx D x x
2 21 5 5 1 x A A1 8
1 5 5 ( 3 ) (7 4 ) ,5 5
x i i B D B D i B D
1 2 1
10 0 2 2
5x A A D A
Example Ex. Find Sol.
4 3 2
2 2 2
3.
( 1)( 1)
x x x x
I dxx x
4 3 21 2 1 1 2 2
2 2 2 2 2 2
3
( 1)( 1) 1 1 1 ( 1)
A A B x D B x Dx x x x
x x x x x x
1 2 1 1 2 2
1 1, , 1, 1, 1, 1
2 2 A A B D B D
22 2
1 1 1 1ln | 1| ln | 1| ln( 1) arctan
2 2 2 ( 1)
xI x x x x dx
x
2
2 2
1 1 1 1ln | | arctan .
2 1 2 2( 1)
x xx C
x x
Homework 18 Section 7.2: 18, 24, 42, 44
Section 7.3: 5, 6, 24, 27